§ 3. Integration of Differential Forms
209
The crucial point is the following lemma, in which K(r) = K(0, r).
Lemma 1. Let U be an open subset of R
n containing 0 and ψ : U −→ R
n a
C
1 map such that ψ(0) = 0, ψ
(0) = 1. Given a number q such that 0 < q < 1,
let r be a number > 0 such that K(r) ⊂ U and
|x| < r =⇒ ⇒ψ
(x) − 1 < q .
(10.11)
Then
K [(1 − q)r] ⊂ ψ [K(r)] ⊂ K [(1 + q)r] .
(10.12)
The derivative ψ
(x) being a continuous function of x equal to 1 at x = 0,
the existence of r for given q > 0 is obvious. Having said that, suppose
x ∈ K(r), so that tx ∈ K(r) for 0 ≤ t ≤ 1. The derivative of t → ψ(tx) is
ψ
(tx)x ; as ψ(0) = 0, ψ(x) =
ψ
(tx)xdt, and so
ψ(x) − x =
[ψ
(tx) − 1] x.dt ,
where integration is over (0, 1). Since, by (11),
|ψ
(tx)x − x| ≤ ≤ψ
(tx) − 1.|x| ≤ ≤ψ
(tx) − 1r ≤ qr
|ψ(x) − x| ≤ qr and so
|ψ(x)| ≤ |x| + r ≤ (1 + q)r ,
which proves the right half of (12).
To prove the other less easy inequality, imitating the proof of the local
inversion theorem is a possibility. It all amounts to showing that, for all
ζ ∈ K[(1 − q)r], there exists z ∈ K(r) such that ψ(z) = ζ. For this, set
ψ(z) = z + p(z). So p
(z) = ψ
(z) − 1 ; by (11),
|z| ≤ r =⇒ |p(z)| ≤ q|z| .
(10.13)
Then construct a sequence of points
z 1 = ζ , z 2 = ζ − p (z 1 ) , z 3 = ζ − p (z 2 ) , . . .
as in Chap. III, § 5, Theorem 24, whose proof we follow (except that, lack of
foresight, q = 1/2 was chosen in Chapter III). We must make sure that the
construction continues without obstruction, i.e. that z 1 ∈ K(r) – obvious –
and that z n ∈ K(r) implies z n+1 ∈ K(r) . Now, by (13),
|z n+1 | ≤ |ζ| + |p(z n )| ≤ (1 − q)r + q|z n | ≤ r .